The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $102$ years; the standard deviation is $9.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living longer than $92.4$ years.
Solution: $102$ $92.4$ $111.6$ $82.8$ $121.2$ $73.2$ $130.8$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $102$ years. We know the standard deviation is $9.6$ years, so one standard deviation below the mean is $92.4$ years and one standard deviation above the mean is $111.6$ years. Two standard deviations below the mean is $82.8$ years and two standard deviations above the mean is $121.2$ years. Three standard deviations below the mean is $73.2$ years and three standard deviations above the mean is $130.8$ years. We are interested in the probability of a turtle living longer than $92.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the turtles will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $92.4$ years and the other half $({16\%})$ will live longer than $111.6$ years. The probability of a particular turtle living longer than $92.4$ years is ${68\%} + {16\%}$, or $84\%$.